\documentclass[handout]{slide}



\renewcommand{\mytitle}{第九章\quad 多元函数微分法及其应用 }
\renewcommand{\mysubtitle}{第四节\quad 多元复合函数的求导法则}
\graphicspath{ {./images/} }
\begin{document}


\begin{frame}{多元复合函数的求导法则}
本节要将一元函数微分学中复合函数的求导法则推广到多元复合函数的情形。多元复合函数的求导法则在多元函数微分学中也起着重要作用。

~

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下面按照多元复合函数不同的复合情形，分三种情形讨论：一元函数与多元函数复合、多元函数与多元函数复合、其他情形。
\end{frame}


\begin{frame}{一元函数与多元函数复合的情形}
  \pause
  \begin{theorem*}
  如果函数 $u=\varphi(t)$ 及 $v=\psi(t)$ 都在点 $t$ 可导， 函数 $z=f(u, v)$ 在对应点 $(u, v)$ 具有连续偏导数， 那么复合函数 $z=f[\varphi(t), \psi(t)]$ 在点 $t$ 可导， 且有
  \[\tag{4-1}
    \frac{\mathrm{d} z}{\mathrm{~d} t}=\frac{\partial z}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial z}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} t}.
\]
\end{theorem*}

\pause
用同样的方法， 可把定理推广到复合函数的中间变量多于两个的情形。 
\pause
例如， 设 $z=f(u, v, w), u=\varphi(t), v=\psi(t), w=\omega(t)$ 复合而得复合函数
\[
z=f[\varphi(t), \psi(t), \omega(t)],
\]
则在与定理相类似的条件下， 这个复合函数在点 $t$ 可导， 且其导数可用下列公式计算：
\[\tag{4-2}
  \frac{\mathrm{d} z}{\mathrm{~d} t}=\frac{\partial z}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial z}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} t}+\frac{\partial z}{\partial w} \frac{\mathrm{d} w}{\mathrm{~d} t}.
\]
\pause
公式 (4-1) 及 (4-2) 中的导数 $\frac{\mathrm{d} z}{\mathrm{~d} t}$ 称为\emph{全导数}。
\end{frame}



\begin{frame}


\begin{proof}
  设 $t$ 获得增量 $\Delta t$, 这时 $u=\varphi(t), v=\psi(t)$ 的对应增量为 $\Delta u, \Delta v$, 由此， 函数 $z=$ $f(u, v)$ 相应地获得增量 $\Delta z$. 按假定， 函数 $z=f(u, v)$ 在点 $(u, v)$ 具有连续偏导数， 这时函数的全增量 $\Delta z$ 可表示为
  \[
  \Delta z=\frac{\partial z}{\partial u} \Delta u+\frac{\partial z}{\partial v} \Delta v+\varepsilon_{1} \Delta u+\varepsilon_{2} \Delta v
\]
这里， 当 $\Delta u \rightarrow 0, \Delta v \rightarrow 0$ 时， $\varepsilon_{1} \rightarrow 0, \varepsilon_{2} \rightarrow 0$. 在偏导数连续的条件下， 上述公式成立的证明参见本章第三节定理 2 的证明。

将上式两端各除以 $\Delta t$, 得
\[
\frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial u} \frac{\Delta u}{\Delta t}+\frac{\partial z}{\partial v} \frac{\Delta v}{\Delta t}+\varepsilon_{1} \frac{\Delta u}{\Delta t}+\varepsilon_{2} \frac{\Delta v}{\Delta t}
\]
因为当 $\Delta t \rightarrow 0$ 时， $\Delta u \rightarrow 0, \Delta v \rightarrow 0, \frac{\Delta u}{\Delta t} \rightarrow \frac{\mathrm{d} u}{\mathrm{~d} t}, \frac{\Delta v}{\Delta t} \rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}$, 所以
\[
  \lim _{\Delta t \rightarrow 0} \frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial z}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} t}.
\]
这就证明了复合函数 $z=f[\varphi(t), \psi(t)]$ 在点 $t$ 可导， 且其导数可用公式 (4-1) 计算。证毕。
\end{proof}

\end{frame}



\begin{frame}{多元函数与多元函数复合的情形}

  \begin{theorem*}
  如果函数 $u=\varphi(x, y)$ 及 $v=\psi(x, y)$ 都在点 $(x, y)$ 具有对 $x$ 及对 $y$ 的偏导数， 函数 $z=f(u, v)$ 在对应点 $(u, v)$ 具有连续偏导数， 那么复合函数 $z=f[\varphi(x, y), \psi(x, y)]$ 在点 $(x, y)$ 的两个偏导数都存在， 且有
  \begin{align}
    \frac{\partial z}{\partial x}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}
    \tag{4-3} \\
    \frac{\partial z}{\partial y}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}
    \tag{4-4}
\end{align}\end{theorem*}

\pause
事实上， 这里求 $\frac{\partial z}{\partial x}$ 时， 将 $y$ 看做常量， 因此 $u=\varphi(x, y)$ 及 $v=\psi(x, y)$ 仍可看做一元函数而应用定理 1. 
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但由于复合函数 $z=f[\varphi(x, y), \psi(x, y)]$ 以及 $u=\varphi(x, y)$ 和 $v=$ $\psi(x, y)$ 都是 $x, y$ 的二元函数，所以应把 (4-1) 式中的 $\mathrm{d}$ 改为 $\partial$, 再把 $t$ 换成 $x$, 这样便由 (4-1) 式得 (4-3) 式。 
\pause
同理由 (4-1) 式可得 (4-4) 式。

\end{frame}


\begin{frame}
  类似地， 设 $u=\varphi(x, y), v=\psi(x, y)$ 及 $w=\omega(x, y)$ 都在点 $(x, y)$ 具有对 $x$ 及对 $y$ 的偏导数， 函数 $z=f(u, v, w)$ 在对应点 $(u, v, w)$ 具有连续偏导数， 则复合函数
\[
z=f[\varphi(x, y), \psi(x, y), \omega(x, y)]
\]
在点 $(x, y)$ 的两个偏导数都存在， 且可用下列公式计算：
\begin{align}
  \frac{\partial z}{\partial x}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}+\frac{\partial z}{\partial w} \frac{\partial w}{\partial x} \tag{4-5} \\
  \frac{\partial z}{\partial y}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}+\frac{\partial z}{\partial w} \frac{\partial w}{\partial y} \tag{4-6}
\end{align}
\end{frame}


\begin{frame}{其他情形}


\begin{theorem*}
  如果函数 $u=\varphi(x, y)$ 在点 $(x, y)$ 具有对 $x$ 及对 $y$ 的偏导数， 函数 $v=\psi(y)$
在点 $y$ 可导， 函数 $z=f(u, v)$ 在对应点 $(u, v)$ 具有连续偏导数， 那么复合函数 $z=f[\varphi(x, y)$, $\psi(y)]$ 在点 $(x, y)$ 的两个偏导数都存在， 且有
\begin{align}
  \frac{\partial z}{\partial x}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} \tag{4-7} \\
  \frac{\partial z}{\partial y}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} y}
  \tag{4-8}
\end{align}
\end{theorem*}
\pause
  上述情形实际上是情形 2 （多元函数与多元函数复合的情形） 的一种特例， 即在情形 2 中，如变量 $v$ 与 $x$ 无关， 从而 $\frac{\partial v}{\partial x}=0$; 在 $v$ 对 $y$ 求导时， 由于 $v=\psi(y)$ 是一元函数， 故 $\frac{\partial v}{\partial y}$ 换成了 $\frac{\mathrm{d} v}{\mathrm{~d} y}$, 这就得出上述结果。

\end{frame}


\begin{frame}


在情形 3 中， 还会遇到这样的情形：复合函数的某些中间变量本身又是复合函数的自变量。例如， 设 $z=f(u, x, y)$ 具有连续偏导数， 而 $u=\varphi(x, y)$ 具有偏导数， 则复合函数 $z=f[\varphi(x, y), x, y]$ 可看做情形 2 （多元函数与多元函数复合的情形）中当 $v=x, w=y$ 的特殊情形。 
\pause
因此
\[
\frac{\partial v}{\partial x}=1, \quad \frac{\partial w}{\partial x}=0, \quad \frac{\partial v}{\partial y}=0, \quad \frac{\partial w}{\partial y}=1
\]
从而复合函数 $z=f[\varphi(x, y), x, y]$ 具有对自变量 $x$ 及 $y$ 的偏导数， 且由公式 (4-5)、 (4-6) 得
\[
\frac{\partial z}{\partial x}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial f}{\partial x}, \quad \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial f}{\partial y}
\]

~

\pause
\begin{warning}
这里 $\frac{\partial z}{\partial x}$ 与 $\frac{\partial f}{\partial x}$ 是不同的， $\frac{\partial z}{\partial x}$ 是把复合函数 $z=f[\varphi(x, y), x, y]$ 中的 $y$ 看做不变而对 $x$ 的偏导数， $\frac{\partial f}{\partial x}$ 是把 $f(u, x, y)$ 中的 $u$ 及 $y$ 看做不变而对 $x$ 的偏导数。 $\frac{\partial z}{\partial y}$ 与 $\frac{\partial f}{\partial y}$ 也有类似的区别。
\end{warning}

\end{frame}


\begin{frame}
\begin{example}
  设 $z=\mathrm{e}^{u} \sin v$, 而 $u=x y, v=x+y$. 求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$.
\end{example}
\pause
\begin{solution} 我们有
  \begin{align*}
    \frac{\partial z}{\partial x}&= \frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}=\mathrm{e}^{u} \sin v \cdot y+\mathrm{e}^{u} \cos v \cdot 1=\mathrm{e}^{x y}[y \sin (x+y)+\cos (x+y)], \\
    \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}=\mathrm{e}^{u} \sin v \cdot x+\mathrm{e}^{u} \cos v \cdot 1=\mathrm{e}^{x y}[x \sin (x+y)+\cos (x+y)].
      \end{align*}
    \end{solution}

\end{frame}


\begin{frame}

    \begin{example}
    设 $u=f(x, y, z)=\mathrm{e}^{x^{2}+y^{2}+z^{2}}$, 而 $z=x^{2} \sin y$. 
    求复合后的函数$u(x,y)$的偏导数 $\frac{\partial u}{\partial x}$ 和 $\frac{\partial u}{\partial y}$.
  \end{example}
\pause
  \begin{solution}
    我们有
    \begin{align*}
      \frac{\partial u}{\partial x}= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial x} &=2 x \mathrm{e}^{x^{2}+y^{2}+z^{2}}+2 z \mathrm{e}^{x^{2}+y^{2}+z^{2}} \cdot 2 x \sin y\\
      &= 2 x\left(1+2 x^{2} \sin ^{2} y\right) \mathrm{e}^{x^{2}+y^{2}+x^{4} \sin ^{2} y}, \\
      \frac{\partial u}{\partial y}= \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial y}&= 2 y \mathrm{e}^{x^{2}+y^{2}+z^{2}}+2 z \mathrm{e}^{x^{2}+y^{2}+z^{2}} \cdot x^{2} \cos y \\
      &= 2\left(y+x^{4} \sin y \cos y\right) \mathrm{e}^{x^{2}+y^{2}+x^{4} \sin ^{2} y}.
            \end{align*}
          \end{solution}
\end{frame}


\begin{frame}


  \begin{example}
    设 $z=f(u, v, t)=u v+\sin t$, 而 $u=\mathrm{e}^{t}, v=\cos t$. 求全导数 $\frac{\mathrm{d} z}{\mathrm{~d} t}$.
  \end{example}
  \pause
  \begin{solution}
  我们有
  \[
    \begin{aligned}
    \frac{\mathrm{d} z}{\mathrm{~d} t} & =\frac{\partial f}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial f}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} t}+\frac{\partial f}{\partial t}=v \mathrm{e}^{t}-u \sin t+\cos t \\
    & =\mathrm{e}^{t} \cos t-\mathrm{e}^{t} \sin t+\cos t=\mathrm{e}^{t}(\cos t-\sin t)+\cos t.
\end{aligned}
\]
\end{solution}
\end{frame}

\begin{frame}
  \begin{example}
   设 $w=f(x+y+z, x y z), f$ 具有二阶连续偏导数， 求 $\frac{\partial w}{\partial x}$ 及 $\frac{\partial^{2} w}{\partial x \partial z}$.
  \end{example}
\pause
 \begin{solution}
  令 $u=x+y+z, v=x y z$, 则 $w=f(u, v)$.
 因所给函数由 $w=f(u, v)$ 及 $u=x+y+z, v=x y z$ 复合而成， 根据复合函数求导法则， 有
  \[
     \begin{aligned}
       \frac{\partial w}{\partial x}&= \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \frac{\partial v}{\partial x}=f_{u}+y z f_{v}, \\
       \frac{\partial^{2} w}{\partial x \partial z}&= \frac{\partial}{\partial z}\left(f_{u}+y z f_{v}\right)=\frac{\partial f_{u}}{\partial z}+y f_{v}+y z \frac{\partial f_{v}}{\partial z} .
      \end{aligned}
     \]
     \pause
    求 $\frac{\partial f_{u}}{\partial z}$ 及 $\frac{\partial f_{v}}{\partial z}$ 时， 应注意 $f_{u}(u, v)$ 及 $f_{v}(u, v)$ 中 $u, v$ 是中间变量， 根据复合函数求导法则， 有
   \[
      \begin{aligned}
        \frac{\partial f_{u}}{\partial z}&= \frac{\partial f_{u}}{\partial u} \frac{\partial u}{\partial z}+\frac{\partial f_{u}}{\partial v} \frac{\partial v}{\partial z}=f_{u u}+x y f_{u v}, \\
        \frac{\partial f_{v}}{\partial z}&= \frac{\partial f_{v}}{\partial u} \frac{\partial u}{\partial z}+\frac{\partial f_{v}}{\partial v} \frac{\partial v}{\partial z}=f_{v u}+x y f_{v v},
       \end{aligned}
      \]
     于是
    \[
     \frac{\partial^{2} w}{\partial x \partial z}=f_{u u}+x y f_{u v}+y f_{v}+y z f_{v u}+x y^{2} z f_{v v}=f_{u u}+y(x+z) f_{u v}+x y^{2} z f_{v v}+y f_{v} .
    \]
   \end{solution}
 \end{frame}

 \begin{frame}
  有时，为表达简便起见，引人以下记号：
  \[
  f_{1}^{\prime}(u, v)=f_{u}(u, v), \quad f_{2}^{\prime}(u, v)=f_{v}(u, v), \quad f_{12}^{\prime \prime}(u, v)=f_{u v}(u, v),
\]
这里，下标 $1$ 表示对第一个变量 $u$ 求偏导数，下标 $2$ 表示对第二个变量 $v$ 求偏导数。 同理有 $f_{11}^{\prime \prime}, f_{22}^{\prime \prime}, f_{21}^{\prime \prime}$ 等。 利用这种记号， 上例的结果可表示成
\[
  \begin{aligned}
    \frac{\partial w}{\partial x}&= f_{1}^{\prime}+y z f_{2}^{\prime}, \\
    \frac{\partial^{2} w}{\partial x \partial z}&= f_{11}^{\prime \prime}+y(x+z) f_{12}^{\prime \prime}+x y^{2} z f_{22}^{\prime \prime}+y f_{2}^{\prime} .
\end{aligned}
\]
\end{frame}

\begin{frame}
\begin{example}
  设 $u=f(x, y)$ 的所有二阶偏导数连续，把下列表达式转换为极坐标系中的形式：

(1) $\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}$; 
\hskip 12em
(2) $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}$.
\end{example}
\pause
由直角坐标与极坐标间的关系式
\[
  x=\rho\cos\theta, \quad y=\rho\sin\theta,
\]
我们可把$u=f(x,y)$换成极坐标 $\rho$ 及 $\theta$ 的函数：
\[
u=f(x, y)=f(\rho \cos \theta, \rho \sin \theta)=F(\rho, \theta) .
\]
所谓把
\[
  \frac{\partial u}{\partial x},\quad \frac{\partial u}{\partial y}, \quad \frac{\partial^{2} u}{\partial x^{2}},\quad \frac{\partial^{2} u}{\partial y^{2}}
\]
转换为极坐标中的形式就是指用 $\rho, \theta$ 及函数 $u=F(\rho, \theta)$ 对 $\rho, \theta$ 的（高阶）偏导数来表达上述函数复合上极坐标方程后得到的函数
\[
  \frac{\partial u}{\partial x}(\rho\cos\theta, \rho\sin\theta),\quad \frac{\partial u}{\partial y}(\rho\cos\theta, \rho\sin\theta), \quad \frac{\partial^2 u}{\partial x^2}(\rho\cos\theta, \rho\sin\theta), \quad \frac{\partial^2 u}{\partial y^2}(\rho\cos\theta, \rho\sin\theta).
\]
\end{frame}

\begin{frame}

\begin{solution}[解法一]
  为了将式子 $\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}$ 及 $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}$ 转化为极坐标的形式，
  我们要求出 $u=f(x, y)$ 的偏导数 $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial^{2} u}{\partial x^{2}}$ 及 $\frac{\partial^{2} u}{\partial y^{2}}$. 
  这里，我们把 $u=f(x, y)$ 看做由 $u=$ $F(\rho, \theta)$ 及
\[
\rho=\sqrt{x^{2}+y^{2}}, \quad \theta=\arctan \frac{y}{x} 
\]
复合而成%
\footnote{当点 $P(x, y)$ 在第一、四像限时， 规定 $\theta$ 的取值范围为 $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, 则
$
\theta=\arctan \frac{y}{x};
$
当点 $P(x, y)$ 在第二、三象限时， 规定 $\theta$ 的取值范围为 $\frac{\pi}{2}<\theta<\frac{3}{2} \pi$, 则
$
\theta=\arctan \frac{y}{x}+\pi.
$
此时以下推导仍成立。没有包含进来的边界上的点可以试着连续延拓过去。}。
应用复合函数求导法则， 得
\[
  \begin{aligned}
  & \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \rho} \frac{\partial \rho}{\partial x}+\frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x}=\frac{\partial u}{\partial \rho} \frac{x}{\rho}-\frac{\partial u}{\partial \theta} \frac{y}{\rho^{2}}=\frac{\partial u}{\partial \rho} \cos \theta-\frac{\partial u}{\partial \theta} \frac{\sin \theta}{\rho} \\
& \frac{\partial u}{\partial y}=\frac{\partial u}{\partial \rho} \frac{\partial \rho}{\partial y}+\frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial y}=\frac{\partial u}{\partial \rho} \frac{y}{\rho}+\frac{\partial u}{\partial \theta} \frac{x}{\rho^{2}}=\frac{\partial u}{\partial \rho} \sin \theta+\frac{\partial u}{\partial \theta} \frac{\cos \theta}{\rho}
\end{aligned}
\]
两式平方后相加， 得
\[
\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial \rho}\right)^{2}+\frac{1}{\rho^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2}
\]
\end{solution}
\end{frame}

\begin{frame}

  \begin{solution}[解法一 (续)]
再求二阶偏导数， 得
\[
  \begin{aligned}
  \frac{\partial^{2} u}{\partial x^{2}} & =\frac{\partial}{\partial \rho}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial \rho}{\partial x}+\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x} \\
& =\left[\frac{\partial}{\partial \rho}\left(\frac{\partial u}{\partial \rho} \cos \theta-\frac{\partial u}{\partial \theta} \frac{\sin \theta}{\rho}\right)\right] \cdot \cos \theta-\left[\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial \rho} \cos \theta-\frac{\partial u}{\partial \theta} \frac{\sin \theta}{\rho}\right)\right] \cdot \frac{\sin \theta}{\rho} \\
& =\frac{\partial^{2} u}{\partial \rho^{2}} \cos ^{2} \theta-\frac{\partial^{2} u}{\partial \rho \partial \theta} \frac{\sin 2 \theta}{\rho}+\frac{\partial^{2} u}{\partial \theta^{2}} \frac{\sin ^{2} \theta}{\rho^{2}}+\frac{\partial u}{\partial \theta} \frac{\sin 2 \theta}{\rho^{2}}+\frac{\partial u}{\partial \rho} \frac{\sin ^{2} \theta}{\rho}.
\end{aligned}
\]
同理可得
\[
  \frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial \rho^{2}} \sin ^{2} \theta+\frac{\partial^{2} u}{\partial \rho \partial \theta} \frac{\sin 2 \theta}{\rho}+\frac{\partial^{2} u}{\partial \theta^{2}} \frac{\cos ^{2} \theta}{\rho^{2}}-\frac{\partial u}{\partial \theta} \frac{\sin 2 \theta}{\rho^{2}}+\frac{\partial u}{\partial \rho} \frac{\cos ^{2} \theta}{\rho}.
\]
两式相加， 得
\[
  \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{2} u}{\partial \rho^{2}}+\frac{1}{\rho} \frac{\partial u}{\partial \rho}+\frac{1}{\rho^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=\frac{1}{\rho^{2}}\left[\rho \frac{\partial}{\partial \rho}\left(\rho \frac{\partial u}{\partial \rho}\right)+\frac{\partial^{2} u}{\partial \theta^{2}}\right].
\]
\end{solution}
\end{frame}


\begin{frame}
\begin{solution}[解法二]
  设$f(x,y)$与极坐标方程$x=\rho \cos\theta, y=\rho\sin \theta$复合后得到的函数为
    $u(\rho,\theta)$, 即
  \[
    u(\rho,\theta)=f(\rho\cos\theta, \rho\sin \theta).
  \]
  那么我们有
  \begin{align*}
    \frac{\partial u}{\partial \rho}&= \cos\theta\cdot \frac{\partial f}{\partial x} + \sin\theta\cdot \frac{\partial f}{\partial y}, \\
    \frac{\partial u}{\partial \theta} &= -\rho\sin\theta\cdot \frac{\partial f}{\partial x} + \rho\cos\theta\cdot \frac{\partial f}{\partial y}.
  \end{align*}
  视$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$为未知应用Cramer法则解此方程组得
      \begin{align*}
        \tag{$*$}
        \frac{\partial f}{\partial x}   &=\cos \theta\cdot \frac{\partial u}{\partial \rho}  - \frac{\sin \theta}{\rho} \cdot \frac{\partial u}{\partial \theta} ,\\
        \tag{$**$}
        \frac{\partial f}{\partial y}   &=\sin \theta\cdot\frac{\partial u}{\partial \rho}   + \frac{\cos \theta}{\rho}\cdot \frac{\partial u}{\partial \theta} .
      \end{align*}
      ($*$)和($**$)两式右边就分别为$\frac{\partial f}{\partial x}$和$\frac{\partial f}{\partial x}$的极坐标形式。
    这样容易算出
    \[
      \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 = 
      \left( \frac{\partial u}{\partial \rho} \right)^2+ \frac{1}{\rho^2}\left( \frac{\partial u}{\partial \theta} \right)^2.
    \]
  \end{solution}
\end{frame}


\begin{frame}
  \begin{solution}[解法二 (续)]
    分别用$\frac{\partial f}{\partial x}$ 和($*$) 右边的式子替换 ($*$)式中$f(x,y)$和$u(\rho,\theta)$的角色可知
    \[
      \begin{aligned}
        \frac{\partial^2 f}{\partial x^2}
        &= \cos \theta \cdot \frac{\partial}{\partial \rho} \left( \cos\theta \cdot \frac{\partial u}{\partial \rho} -\frac{\sin \theta}{\rho}\cdot \frac{\partial u}{\partial \theta}  \right) 
        - \frac{\sin \theta}{\rho} \frac{\partial }{\partial \theta} \left( \cos\theta \cdot \frac{\partial u}{\partial \rho} -\frac{\sin \theta}{\rho}\cdot \frac{\partial u}{\partial \theta}  \right) \\
        &= \cos \theta \cdot \left( \cos\theta \cdot \frac{\partial^2 u}{\partial \rho^2} - \left( -\frac{\sin \theta}{\rho^2}\frac{\partial u}{\partial \theta} + \frac{\sin \theta}{\rho}\frac{\partial^2 u}{\partial \rho \partial \theta}  \right) \right) \\
        & \quad - \frac{\sin \theta}{\rho} \left( \left( -\sin \theta\cdot \frac{\partial u}{\partial \rho} + \cos \theta \frac{\partial^2 u}{\partial \theta \partial \rho} \right) - \left( \frac{\cos \theta}{\rho} \frac{\partial u}{\partial \theta} + \frac{\sin \theta}{\rho} \frac{\partial^2 u}{\partial \theta^2} \right)  \right) \\
        &= \frac{\partial^{2} u}{\partial \rho^{2}} \cos ^{2} \theta-\frac{\partial^{2} u}{\partial \rho \partial \theta} \frac{\sin 2 \theta}{\rho}+\frac{\partial^{2} u}{\partial \theta^{2}} \frac{\sin ^{2} \theta}{\rho^{2}}+\frac{\partial u}{\partial \theta} \frac{\sin 2 \theta}{\rho^{2}}+\frac{\partial u}{\partial \rho} \frac{\sin ^{2} \theta}{\rho}.
      \end{aligned}
  \]
    类似地，
    \[
      \frac{\partial^2 f}{\partial y^2} =\frac{\partial^{2} u}{\partial \rho^{2}} \sin ^{2} \theta+\frac{\partial^{2} u}{\partial \rho \partial \theta} \frac{\sin 2 \theta}{\rho}+\frac{\partial^{2} u}{\partial \theta^{2}} \frac{\cos ^{2} \theta}{\rho^{2}}-\frac{\partial u}{\partial \theta} \frac{\sin 2 \theta}{\rho^{2}}+\frac{\partial u}{\partial \rho} \frac{\cos ^{2} \theta}{\rho}.
    \]
    上面两式相加即可得 
    \[
      \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{1}{\rho^{2}}\left[\rho \frac{\partial}{\partial \rho}\left(\rho \frac{\partial u}{\partial \rho}\right)+\frac{\partial^{2} u}{\partial \theta^{2}}\right].
\]
  \end{solution}
\end{frame}

\begin{frame}
\begin{remark}
  关于未知量$x_1, x_2$的二元线性方程组
\[
\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}=b_{1} \\
a_{21} x_{1}+a_{22} x_{2}=b_{2}
\end{array}\right.
\]
的Cramer法则说的是：若系数行列式
\[
  \begin{vmatrix}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
  \end{vmatrix}\coloneq a_{11}a_{22}-a_{12}a_{21}\neq 0,
\]
那么该线性方程组有唯一解：
\[
x_{1}=\frac{\left|\begin{array}{ll}
b_{1} & a_{12}  \\
b_{2} & a_{22}
\end{array}\right|}{\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|}, \quad x_{2}=\frac{\left|\begin{array}{ll}
a_{11} & b_{1} \\
a_{21} & b_{2}
\end{array}\right|}{\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|}
\]
一般地，由$n$个方程构成的$n$元线性方程组有Cramer法则（这也是行列式这个概念的由来）。
\end{remark}
\end{frame}


\begin{frame}{全微分形式不变性}

  设函数 $z=f(u, v)$ 具有连续偏导数， 则有全微分
\[
\mathrm{d} z=\frac{\partial z}{\partial u} \mathrm{~d} u+\frac{\partial z}{\partial v} \mathrm{~d} v
\]
\pause
如果 $u$ 和 $v$ 又是中间变量， 即 $u=\varphi(x, y), v=\psi(x, y)$, 且这两个函数也具有连续偏导数，那么复合函数
\[
z=f[\varphi(x, y), \psi(x, y)]
\]
的全微分为
\[
  \mathrm{d} z=\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y,
\]
其中 $\frac{\partial z}{\partial x}$ 及 $\frac{\partial z}{\partial y}$ 分别由公式 (4-3) 及 (4-4) 给出。 
\pause
把公式 (4-3) 及 (4-4) 中的 $\frac{\partial z}{\partial x}$ 及 $\frac{\partial z}{\partial y}$ 代人上式， 得
\[
  \begin{aligned}
  \mathrm{d} z & =\left(\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}\right) \mathrm{d} x+\left(\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}\right) \mathrm{d} y \\
  & =\frac{\partial z}{\partial u}\left(\frac{\partial u}{\partial x} \mathrm{~d} x+\frac{\partial u}{\partial y} \mathrm{~d} y\right)+\frac{\partial z}{\partial v}\left(\frac{\partial v}{\partial x} \mathrm{~d} x+\frac{\partial v}{\partial y} \mathrm{~d} y\right)=\frac{\partial z}{\partial u} \mathrm{~d} u+\frac{\partial z}{\partial v} \mathrm{~d} v.
\end{aligned}
\]
\pause
由此可见，无论 $u$ 和 $v$ 是自变量还是中间变量， 函数 $z=f(u, v)$ 的全微分形式是一样的。这个性质叫做\emph{全微分形式不变性}。
\end{frame}


\begin{frame}
  \begin{example}
  利用全微分形式不变性解本节的例 1.
\end{example}
\pause
\begin{solution}
  $\mathrm{d} z=\mathrm{d}\left(\mathrm{e}^{u} \sin v\right)=\mathrm{e}^{u} \sin v \mathrm{~d} u+\mathrm{e}^{u} \cos v \mathrm{~d} v,$
因
\[
\mathrm{d} u=\mathrm{d}(x y)=y \mathrm{~d} x+x \mathrm{~d} y, \mathrm{~d} v=\mathrm{d}(x+y)=\mathrm{d} x+\mathrm{d} y,
\]
代人后归并含 $\mathrm{d} x$ 及 $\mathrm{d} y$ 的项， 得
\[
\mathrm{d} z=\left(\mathrm{e}^{u} \sin v \cdot y+\mathrm{e}^{u} \cos v\right) \mathrm{d} x+\left(\mathrm{e}^{u} \sin v \cdot x+\mathrm{e}^{u} \cos v\right) \mathrm{d} y,
\]
即
\[
\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y=\mathrm{e}^{x y}[y \sin (x+y)+\cos (x+y)] \mathrm{d} x+\mathrm{e}^{x y}[x \sin (x+y)+\cos (x+y)] \mathrm{d} y .
\]
比较上式两端的 $\mathrm{d} x$ 和 $\mathrm{d} y$ 的系数， 就同时得到两个偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$, 它们与例 1 的结果一样。
\end{solution}
\end{frame}

\end{document}
